Question

plz help me 11th pcbm Math problem plz
Test the continuity of
$f(x) = \frac{y {}^{x} - 2 {}^{ x+ 1} + 1}{1 - \cos2x }$
for x=/ 0
$= ( \frac{ log2}{2}) {}^{2}$
for x=0 at x=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Appropriate function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x)-\begin{cases} &\sf{\dfrac{ {4}^{x} - {2}^{ x+ 1} + 1}{1 - \cos2x } \: when \: x \: \ne \: 0} \\ \\ &\sf{\bigg(\dfrac{ log2}{2}\bigg)^{2} \: when \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered}$

We know that,

A function f(x) is said to be continuous at x = a, iff

$\boxed{ \quad \sf{ \:\displaystyle\lim_{x \to a} \: \bf \: f(x) = f(a) \quad}}$

Now,

$\rm :\longmapsto\:f(0) = \bigg(\dfrac{ log2}{2}\bigg) ^{2} - - - (1)$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \: f(x)$

$\rm \:  =  \:  \:\displaystyle\lim_{x \to 0}\frac{{4}^{x} -{2}^{ x+ 1} + 1}{1 - \cos2x }$

If we substitute directly x = 0, we get indeterminant form.

So, the above can be rewritten as

$\rm \:  =  \:  \:\displaystyle\lim_{x \to 0}\frac{{2}^{2x} -2.{2}^{x} + 1}{1 - \cos2x }$

We know that

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

and

$\boxed{ \bf{ \:1 - cos2x = 2 {sin}^{2}x}}$

So, using this, we get

$\rm \:  =  \:  \:\displaystyle\lim_{x \to 0}\dfrac{ {( {2}^{x} - 1)}^{2} }{2 {sin}^{2} x}$

$\rm \:  =  \:  \:\displaystyle\lim_{x \to 0}\dfrac{ {( {2}^{x} - 1)}( {2}^{x} - 1)}{2 {sin}x \times sinx}$

$\rm \:  =  \:  \:\displaystyle\lim_{x \to 0}\dfrac{\dfrac{ {2}^{x} - 1}{x} \times \dfrac{ {2}^{x} - 1}{x} }{2 \times \dfrac{sinx}{x} \times \dfrac{sinx}{x} }$

We know,

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{sinx}{x} = 1}}$

and

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{ {a}^{x} - 1}{x} = loga}}$

Using these results, we get

$\rm \:  =  \:  \:\dfrac{ log(2) \times log(2) }{2 \times 1 \times 1}$

$\rm \:  =  \:  \:\dfrac{ {(log2)}^{2} }{2}$

So, we concluded that

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}f(x) \: \ne \: f(0)$

• Hence, f(x) is not continuous at x = 0.

Additional Information :-

1. If f and g are continuous function at x = a, then

• f + g is also continuous at x = a

• f - g is also continuous at x = a

• f × g is also continuous at x = a

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{tanx}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{tan^{ - 1} x}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{sin^{ - 1} x}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{log(1 + x)}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \bf \: \frac{ {e}^{x} - 1}{x} = 1}}$